The steps i took into doing these problems did very well for me
The general solution of a differential equation of the form ay'' + by' + cy is given by Ae^mx + Be^nx; where m and n are the root of the quadratic equation ax^2 + bx + c
The general solution of <span>y'' + 4y' is given by </span><span>Ae^mx + Be^nx; where m and n are the root of the quadratic equation x^2 + 4x = 0
x(x + 4) = 0
x = 0 or x = -4
Therefore, the general solution is Ae^(-4x) + Be^0x = Ae^(</span>-4x) + B
There are many but I can name some. 1/6 and 5/6. Is that the question you are asking?
Answer:
![6 \frac{1}{6}](https://tex.z-dn.net/?f=6%20%5Cfrac%7B1%7D%7B6%7D%20)
Step-by-step explanation:
![37 \div 6 = 6 \: remainder \: 1 \\ = 6 \frac{1}{6}](https://tex.z-dn.net/?f=37%20%5Cdiv%206%20%3D%206%20%5C%3A%20remainder%20%5C%3A%201%20%5C%5C%20%20%3D%206%20%5Cfrac%7B1%7D%7B6%7D%20)